Nice olympiad inequality

Question

Let's go for an olympiad inequality :

let $a,b,c>0$ then we have : $$\sum_{cyc}\frac{ab}{a+b}\geq \frac{3\sqrt{3}}{2}\sqrt{\frac{abc}{a+b+c}}$$

My proof : $$\sum_{cyc}\frac{ab}{a+b}=\frac{(a^2 b^2 + 3 a^2 b c + a^2 c^2 + 3 a b^2 c + 3 a b c^2 + b^2 c^2)}{(a+b)(b+c)(c+a)}=\frac{(ab + ac + bc)^2+abc(a+b+c)}{(ab+bc+ca)(a+b+c)-abc}$$

So if we use the $uwv's$ method we get :

$$\frac{9v^4+w^3(3u)}{3v^2(3u)-w^3}\geq \frac{3\sqrt{3}}{2}\sqrt{\frac{w^3}{3u}}$$

But I can't conclue by myself so can you help me ?

Answer 1

By your work we need to prove that $f(w^3)\geq0,$ where $$f(w^3)=3v^4+uw^3+\frac{1}{2}(w^3-9uv^2)\sqrt{\frac{w^3}{u}}.$$ We see that $f$ increases, which says that it's enough to prove our inequality for a minimal value of $w^3$, which happens in the following cases.

1. $w^3\rightarrow0^+$.

In this case our inequality is obvious;

2. Two variable are equal.

Since our inequality is homogeneous and symmetric, it's enough to assume $b=c=1$,

which gives $$\frac{5a+1}{a+1}\geq3\sqrt3\sqrt{\frac{a}{a+2}},$$ which is wrong for $a\rightarrow+\infty.$